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interval given ?

with out interval this maximum value diverges to infinity.

wait

1/that ? or, just that ?

lol its
\(\large \color{black}{\begin{align}
\dfrac{1}{x^2-3x+2} \hspace{.33em}\\~\\
\end{align}}\)

well, I can tell you that the limit as x approaches \(\pm\)infinity is going to be 0.

we can plot, I am guessing it is something like
|dw:1435010239497:dw|

none

it is +infinity

but how ?

https://www.desmos.com/calculator

how i prove that with algebra

even the minimum doesn't exist.

with algebra, even with no calc ?

idk calculus

lets c

\(\large \color{black}{\begin{align}
x=2,\ 1 \hspace{.33em}\\~\\
\end{align}}\)

yes

that has a max actually

yes?

https://www.desmos.com/calculator
keep scrollin up

on a certain interval, no doubt (unless it has a domain gap)

i think it attains one max
does not perhaps required any domain to get it or i think

your graph does not show up
see here
https://www.desmos.com/calculator

i don't think there is a max. that is due to an asymptote - two of them, there.

we can find the limit from the right and from the left as x approaches 1 and 2.

http://prntscr.com/7k4q1l
here see that bottom part of the graph

with calculus we can prove it has a max value
but @mathmath333 is looking for a different way

i mean here local max not abolute of course

how to find that local max

oh, but the question is asking for absolute doesn't it?

that is what i would think of course

first let's write that as 1/(x-1)(x-2)

i had thought of absolute too, just not sure about it!

yes the question is asking absolute, i just asked that of curiosity

this function is pretty weird. I wonder what would it model

i mean in real world

you want to then prove there is no max algebraically

yes ^

would that include using ideas from calc but not its techniques?

ok i will try to bear with calculus

\(\large\color{black}{ \displaystyle \frac{1 }{x^2-3x+2} }\)
we need to do only 1 asymptote for this

\(\large\color{black}{ \displaystyle f(0.9)=? }\)

= about 9

f(0.99)=99

try plugging it yourself for close values to 1

these values are called approaching "from the left" because they are a buit less than 1

i got to go

sorry

suppose there exist a max
max(1/(x^2-3x+2)=c
1/x^2-3x+21/c=l
x^2-3x+2-l>0

just some random stuff lol

coming back..

ok i need to go please comment if u found something

let's first consider the domain
\(D_f: \{ x\in \mathbb{R}: x\ne1, x\ne 2 \}\)

that is (-oo, 1)u(1, 2)u(2,oo)

from the graph we knew for 1